Py的函数II

数字漩涡

Py的函数I

<pre><code>if n == 0: print(1) else: a = 1 b = 3 while n &gt;= 2: a,b = b,3*b+2*a+1 n -= 1 print(b % 20132013)</code></pre>

不能递归那就迭代

递归复杂度过高# n = 5 # def solve_it(n): # if n == 0: # return 1 # elif n == 1: # return 3 # else: # return 3*solve_it(n-1) + 2*solve_it(n-2) + 1 # print(solve_it(n) % 20132013) # n = 5自底向上计算就行 def solve_it(n): l = [1,3] if n &lt; 2: return l[n] else: for i in range(2,n+1): answer = 3*l[i-1] + 2*l[i-2] + 1 l.append(answer) return l[-1] print(solve_it(n) % 20132013)

# 简单的数学题
f = [1, 3]
# n =5
for i in range(2,n+1):
    f.append(0)
    f[i] = f[i-1]*3 + f[i-2]*2 +1
print(f[n]%20132013)

自底向上

<pre><code>f0 = 1 f1 = 3 if n == 0: print(1) elif n == 1: print(3) else: for i in range(2, n + 1): fn = (3 * f1 + 2 * f0 + 1) % 20132013 f0 = f1 f1 = fn print(fn)</code></pre>

这个简单

<a href="http://www.gznb.xyz/posts/pythonTip-83-Py的函数I/">题解： http://www.gznb.xyz/posts/pythonTip-83-Py的函数I/</a>

ans = [1, 3]
m=20132013
for i in range(2, n+1):
    r = (((3%m) * (ans[i-1]%m))%m +  ((2%m) * (ans[i-2]%m))%m + 1) % m
    ans.append(r)
print(ans[n])

L=[1,3] for i in range(2,n+1): &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; L.append((3*L[i-1]+2*L[i-2]+1)%20132013) if n==0: &nbsp; &nbsp; print 1 else: &nbsp; &nbsp; print L[-1]

递归不如数列

l=[1,3] for i in range(2,n+1): &nbsp; &nbsp; l.append(0) &nbsp; &nbsp; l[i]=(3*l[i-1]+2*l[i-2]+1)%20132013 print l[n]

enmmmm

<pre style="background-color:#2b2b2b;color:#a9b7c6;font-family:'宋体';font-size:10.5pt;">def F(n):#把递归函数换成for循环
 if n == 0:
 b = 1
 elif n == 1:
 b = 3
 else:
 a = 1
 b = 3
 for i in range(2, n+1):
 c = 2*a+3*b+1
 a = b
 b = c
 return b
print(F(n) % 20132013)
</pre>

将递归换成for循环就可以过了

<pre class="prettyprint lang-py linenums">x,y = 1,3
while n&gt;0:
 x,y,n = y,2*x+3*y+1,n-1
print(pow(x,1,20132013))</pre>

嘿嘿

t1=1 t2=3 tmp=0 for i in range(1,n): &nbsp; &nbsp; tmp=((3*t2)%20132013+(2*t1)%20132013+1)%20132013 &nbsp; &nbsp; t1=t2 &nbsp; &nbsp; t2=tmp if n==0: &nbsp; &nbsp; print t1 elif n==1: &nbsp; &nbsp; print t2 else: &nbsp; &nbsp; print tmp

迭代，没啥难度

与计算Fibonacci的解法是一样的，只是根据模数定理，我们在更新数据的时候，得到新的数据要去取一次模<pre class="prettyprint lang-py linenums">def f(n):
 first = 1
 second = 3
 if n == 0:
 return first
 elif n == 1:
 return second
 else:
 count = 1
 while count != n:
 new_value = 3 * second + 2 * first + 1
 new_value = new_value % 20132013
 first = second
 second = new_value
 count += 1
 	return new_value
#n = 2
print f(n)</pre>

类似Fibonacci的解法

<pre class="prettyprint lang-py linenums">def F(n):
 if n == 0:
 return 1
 elif n == 1:
 return 3
 else:
 return 3*F(n-1)+2*F(n-2)+1
print F(5)%20132013</pre>

def F(n):    if n == 0:        return 1    elif n == 1:        return 3    else:        return 3*F(n-1)+2*F(n-2)+1print F(5)%20132013

def F(n):    if n == 0:        return 1    elif n == 1:        return 3    else:        return 3*F(n-1)+2*F(n-2)+1print F(n)%20132013

这算法怎么优化啊

a,b = 1,3 r = [(a,b)] a,b = b,(3*b+2*a+1)%20132013 while (a,b) not in r: &nbsp;&nbsp;&nbsp; r.append((a,b)) &nbsp;&nbsp;&nbsp; a,b = b,(3*b+2*a+1)%20132013 &nbsp;&nbsp; &nbsp; rbi = r.index((a,b)) r0 = r[:rbi] r = r[rbi:] if n&lt;rbi: &nbsp;&nbsp;&nbsp; print r0[rbi][0] else: &nbsp;&nbsp;&nbsp; print r[(n-rbi)%len(r)][0]

第83题：Py的函数I

信息加密

最小公倍数I

滑雪比赛

反素数(逆向拼写的素数)